Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

Reffer the attached file

Hope it helps :)

From attached file , a= cos5x, b= isin5x c = cos5x and d= isin5x

• Consider option (1)
• a²-b² = cos²5x -(isin5x)²
• a²-b² = cos²5x -(i²)(sin²5x)
• a²-b² = cos²5x - (-1)(sin²5x)
• a²-b² = cos²5x + sin²5x
• Let 5x = y , we know , cos²y + sin²y = 1
• a² - b² = 1
• So , option (1) a²-b² = 1/2 is a false statement

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$A= \displaystyle\begin{pmatrix} \cos \theta & i \sin \theta \: \\ - i\sin \theta & \cos \theta \end{pmatrix}$

So

${A}^{2} = \displaystyle\begin{pmatrix} \cos \theta & i \sin \theta \: \\ - i\sin \theta & \cos \theta \end{pmatrix} \times \displaystyle\begin{pmatrix} \cos \theta & i \sin \theta \: \\ - i\sin \theta & \cos \theta \end{pmatrix}$

$\implies \: {A}^{2} = \displaystyle\begin{pmatrix} {\cos}^{2} \theta - {\sin}^{2} \theta& i \sin \theta \cos \theta + i \sin \theta \cos \theta\: \\ - i \sin \theta \cos \theta - i \sin \theta \cos \theta & {\cos}^{2} \theta - {\sin}^{2} \theta \end{pmatrix}$

$\implies \: {A}^{2} = \displaystyle\begin{pmatrix} \cos2 \theta & i \sin 2\theta \: \\ - i\sin2 \theta & \cos 2\theta \end{pmatrix}$

Similarly

$\implies \: {A}^{5} = \displaystyle\begin{pmatrix} \cos \: 5 \theta & i \sin 5\theta \: \\ - i\sin \: 5 \theta & \cos 5\theta \end{pmatrix}$

Again it is given that

$\: {A}^{5} = \displaystyle\begin{pmatrix} a &b \: \\ c & d \end{pmatrix}$

So

$a = \cos 5 \theta \: \: and \: \: c = - i \sin 5 \theta$

So

${a}^{2} - {c}^{2}$

$= { \cos }^{2} 5 \theta \: - {i}^{2} { \sin }^{2} 5 \theta$

$= { \cos }^{2} 5 \theta \: + { \sin }^{2} 5 \theta \: \: \: ( \because \: {i}^{2} = - 1)$

$= 1$

HENCE THE CORRECT OPTION IS

$\boxed{ {a}^{2} - {c}^{2} = 1}$